# Determine intervals on which s(t) =equation

Determine the intervals on which

$$s(t) = \frac{|t^2-2t - 3|}{t + 1}$$

is continuous.

Hint: Use continuity checklist and check left and right continuity of proposed intervals which include endpoints.

Any help would be appreciated. I'm not sure how to go about solving this one.

Thanks.

• The second one. Everything over (t+1) Commented Jul 7, 2014 at 22:41

$$s(t) = \frac{|t^2-2t-3|}{t+1} = \frac{|(t+1)(t-3)|}{t+1}$$

So $s(t) = |t-3|$ for $t > -1$, $s(t) = t-3$ for $t < -1$ and will be undefined for $t=-1$. Hence it will be continuous on the intervals $(-\infty,-1) \cup (-1, \infty)$.

• Thanks for the help, this is very concise. Do I need to do something with left and right continuity and endpoints? Commented Jul 7, 2014 at 22:44
• You can show that there is a jump discontinuity at $t=-1$, so it isn't possible to redefine $s(-1)$ to make it continuous there.
– MPW
Commented Jul 7, 2014 at 22:47
• A minor point, but $s(t)=\lvert t - 3 \rvert$ for $t > -1$ and $s(t)=(-1)(-(t -3))=t - 3$ for $t < -1$. Commented Jul 7, 2014 at 22:59
• Okay, thanks guys! Commented Jul 7, 2014 at 23:05
• Thanks, Shawn O'Hare, I'll fix it. Commented Jul 7, 2014 at 23:12

The first thing to note is that $s(t)$ is not defined when $t = -1$, so we can only talk about continuity on $\mathbb R \setminus \{-1 \} :=(-\infty, -1) \cup (-1, \infty)$. Now $$s(t)= \frac{\lvert t^2 -2t -3 \rvert}{t + 1} = \frac{\lvert(t+1)(t-3) \rvert}{t+1} = \frac{\lvert t + 1 \rvert}{t + 1} \cdot \lvert t - 3 \rvert.$$ But $$\frac{\lvert t + 1 \rvert}{t + 1} = \begin{cases} 1 & \text{if t > -1,} \\ -1 & \text{if t < -1.} \end{cases}$$ Since $\lvert t - 3 \rvert$ is continuous on all of $\mathbb R$, the only possible source of discontinuity occurs at $t=-1$, which is excluded from the domain of $s(t)$. Hence $s(t)$ is continuous everywhere on its domain, which in this discussion we have taken to be $(-\infty, -1) \cup (-1, \infty)$. The intervals of continuity would be the two open intervals $(-\infty, -1)$ and $(-1, \infty)$.